<Codesignal> Swap Adjacent Bits

Easy

Implement the missing code, denoted by ellipses. You may not modify the pre-existing code.

You're given an arbitrary 32-bit integer n. Take its binary representation, split bits into it in pairs (bit number 0 and 1, bit number 2 and 3, etc.) and swap bits in each pair. Then return the result as a decimal number.

Example

• For n = 13, the output should be
  swapAdjacentBits(n) = 14.

  1310 = 11012 ~> {11}{01}2 ~> 11102 = 1410.

• For n = 74, the output should be
  swapAdjacentBits(n) = 133.

  7410 = 010010102 ~> {01}{00}{10}{10}2 ~> 100001012 = 13310.
  Note the preceding zero written in front of the initial number: since both numbers are 32-bit integers, they have 32 bits in their binary representation. The preceding zeros in other cases don't matter, so they are omitted. Here, however, it does make a difference.

Input/Output

• [execution time limit] 0.5 seconds (cpp)

• [input] integer n

  Guaranteed constraints:
  0 ≤ n < 2^30.

• [output] integer

[C++] Syntax Tips

// Prints help message to the console
// Returns a string
std::string helloWorld(std::string name) {
    std::cout << "This prints to the console when you Run Tests" << std::endl;
    return "Hello, " + name;
}

---

Solution

int swapAdjacentBits(int n) {
  return (n & 0x55555555) << 1 | (n & 0xAAAAAAAA) >> 1;
}